From the mathematical perspective, one can take the magnitude of a cross product: $$ |a\times b|=|a| |b| \sin{\theta}\cdot n, $$ where $\theta$ is the angle between a and b in the plane containing them, and n is the unit vector perpendicular to them.

Does this apply to the cross product of two bra-kets? I became curious from equation 10 on Dr. Berry's paper, where the Berry phase acquired by a state is the double integral of the following term: $$ \langle \nabla n | m\rangle \times \langle m | \nabla n\rangle. $$

So, does the following make sense?

$$ | \langle \nabla n | m\rangle \times \langle m | \nabla n\rangle| = | \langle \nabla n | m\rangle| | \langle m | \nabla n\rangle| \cdot\sin{\theta}, $$


$$ \cos{(\theta)}=\langle \nabla n | m\rangle \cdot\langle m | \nabla n\rangle. $$

To me, it doesn't make sense intuitively because cross products are usually given in tensorial notation in physics (for example: Expressing the magnitude of a cross product in indicial notation). Why is the application of the magnitude above incorrect, or when is it applicable?

  • 4
    $\begingroup$ Note that the cross-product is defined in 3-dimensional space. What is the dimensionality of the Hilbert space of your quantum system? $\endgroup$ Commented Mar 9, 2019 at 23:03
  • 2
    $\begingroup$ @Dan Yand Your comment looks like a fine tight answer to prevent misreadings of Berry, all right. $\endgroup$ Commented Mar 9, 2019 at 23:52
  • $\begingroup$ If @Dan's interpretation of the question is correct then my comment is way off base. :;sigh:: $\endgroup$ Commented Mar 9, 2019 at 23:54
  • 1
    $\begingroup$ @dmckee Well, yes and no... It emphasizes that the Hilbert space as a vector space is aggressively irrelevant here... $\endgroup$ Commented Mar 10, 2019 at 1:01
  • $\begingroup$ Thank you, everyone. @DanYand it appears as if the mathematical definition for the magnitude of the cross product then holds. If someone could write that as an answer I will gladly accept it. As an aside, does this mean that we can use this convention to define the magnitude of Berry curvature? I am having trouble finding such magnitudes in the literature (except in numerical works). $\endgroup$ Commented Mar 10, 2019 at 1:14

2 Answers 2


Usually, when we write $|\cdots\rangle$, the symbol(s) contained inside are just labels used to distinguish different elements of the Hilbert space from each other. That label can be a scalar, vector, matrix, or a list of our favorite arthropod species — but it's still labeling just a single element of the Hilbert space.

However, Berry's paper [1] is using a slightly unconventional notation $|\nabla n\rangle$ in which each component $\nabla_k n$ of $\nabla n$ labels a different element of the Hilbert space, so $|\nabla n\rangle$ is actually an abbreviation for three different kets, namely the kets $|\nabla_k n\rangle$ with $k\in\{1,2,3\}$.

Therefore, in Berry's equation ($7$c), the quantity $v = \langle m|\nabla n\rangle$ is a vector with three components $v_k = \langle m|\nabla_k n\rangle$ with $k\in\{1,2,3\}$. Its complex conjugate is $v^*=\langle\nabla n|m\rangle$. Both $v$ and $v^*$ are ordinary vectors with 3 complex components each. The fact that their components were constructed using bras and kets does not affect the meaning of the cross product $v^*\times v$.

The only unconventional feature of $v^*\times v$ is the fact that the components of the vectors are complex numbers, and part of the question is whether or not a familiar identity for the magnitude of the cross-product generalizes to this case. In particular, the relationship between $|v^*\times v|$ and $v^*\cdot v$ is questioned. To address this, write $v=v_R+iv_I$ where $v_R$ and $v_I$ are the real and imaginary parts of $v$. Then $$ v^*\times v = 2i\,v_R\times v_I \hskip2cm v^*\cdot v = v_R\cdot v_R + v_I\cdot v_I. $$ This shows that the quantity $v^*\times v$ depends on the angle between $v_R$ and $v_I$, but the quantity $v^*\cdot v$ does not. Therefore, the value of $v^*\cdot v$ cannot determine the value of $|v^*\times v|$. This shows that the relationship questioned in the OP cannot hold in general.

As a check, consider the case $v=(1,z,0)$ with $z=\exp(i\phi)$. Then $$ |v^*\times v|=2|\sin\phi| \hskip2cm v^*\cdot v=2 $$ for all $\phi$. This confirms that $v^*\cdot v$ does not determine the value of $|v^*\times v|$.


[1] Berry (1984), "Quantal phase factors accompanying adiabatic changes," https://michaelberryphysics.files.wordpress.com/2013/07/berry120.pdf


Actually, the entire equation is

$$ \langle \nabla n |\times | \nabla n\rangle \rightarrow \sum_{m \neq n} \langle \nabla n | m\rangle \times \langle m | \nabla n\rangle$$

EDIT: It's Stoke's Theorem using a differential 2-form.

Stokes Theorem can be generalized to higher dimensions using differential forms.

The $ m\neq n$ produces the wedge product.

And since $\langle \nabla n | m\rangle$ is the complex conjugate of $\langle m | \nabla n\rangle$ there are only 3 independent quantities.

See Differential Form


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.